Solve for x, y
x=6
y=15
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90x+97y=1995,x+y=21
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
90x+97y=1995
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
90x=-97y+1995
Subtract 97y from both sides of the equation.
x=\frac{1}{90}\left(-97y+1995\right)
Divide both sides by 90.
x=-\frac{97}{90}y+\frac{133}{6}
Multiply \frac{1}{90} times -97y+1995.
-\frac{97}{90}y+\frac{133}{6}+y=21
Substitute -\frac{97y}{90}+\frac{133}{6} for x in the other equation, x+y=21.
-\frac{7}{90}y+\frac{133}{6}=21
Add -\frac{97y}{90} to y.
-\frac{7}{90}y=-\frac{7}{6}
Subtract \frac{133}{6} from both sides of the equation.
y=15
Divide both sides of the equation by -\frac{7}{90}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{97}{90}\times 15+\frac{133}{6}
Substitute 15 for y in x=-\frac{97}{90}y+\frac{133}{6}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{-97+133}{6}
Multiply -\frac{97}{90} times 15.
x=6
Add \frac{133}{6} to -\frac{97}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=6,y=15
The system is now solved.
90x+97y=1995,x+y=21
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}90&97\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1995\\21\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}90&97\\1&1\end{matrix}\right))\left(\begin{matrix}90&97\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}90&97\\1&1\end{matrix}\right))\left(\begin{matrix}1995\\21\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}90&97\\1&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}90&97\\1&1\end{matrix}\right))\left(\begin{matrix}1995\\21\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}90&97\\1&1\end{matrix}\right))\left(\begin{matrix}1995\\21\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{90-97}&-\frac{97}{90-97}\\-\frac{1}{90-97}&\frac{90}{90-97}\end{matrix}\right)\left(\begin{matrix}1995\\21\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{7}&\frac{97}{7}\\\frac{1}{7}&-\frac{90}{7}\end{matrix}\right)\left(\begin{matrix}1995\\21\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{7}\times 1995+\frac{97}{7}\times 21\\\frac{1}{7}\times 1995-\frac{90}{7}\times 21\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\15\end{matrix}\right)
Do the arithmetic.
x=6,y=15
Extract the matrix elements x and y.
90x+97y=1995,x+y=21
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
90x+97y=1995,90x+90y=90\times 21
To make 90x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 90.
90x+97y=1995,90x+90y=1890
Simplify.
90x-90x+97y-90y=1995-1890
Subtract 90x+90y=1890 from 90x+97y=1995 by subtracting like terms on each side of the equal sign.
97y-90y=1995-1890
Add 90x to -90x. Terms 90x and -90x cancel out, leaving an equation with only one variable that can be solved.
7y=1995-1890
Add 97y to -90y.
7y=105
Add 1995 to -1890.
y=15
Divide both sides by 7.
x+15=21
Substitute 15 for y in x+y=21. Because the resulting equation contains only one variable, you can solve for x directly.
x=6
Subtract 15 from both sides of the equation.
x=6,y=15
The system is now solved.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}